Homework3B_jbl5232.xlsx - Joel Luckenbaugh EGEE 430 Homework 3B �/�=_1 〖 〗 �/�≅ 〖 0= 〗_1 〖 15.11 Assumptions/= 〖 〗 〖 〗 Where

# Homework3B_jbl5232.xlsx - Joel Luckenbaugh EGEE 430...

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Joel Luckenbaugh EGEE 430 Homework 3B 3/27/2019 ?[𝑁𝑂]/?𝑡 ?[𝑁]/?𝑡≅
15.11 Assumptions Where k(1f) is 1.82e11*exp(-38370/T(K)) m^3/kmol-s 15.12 k1= 7.6*10^10*(exp(-38000/T(K))) R(u)= 8314.51 Using the following equation, Given: i X(O2)= 0.2 dX(NO)/dt= 6.862567E-06 X(N2)= 0.67 T= 2000 P= 1 k1= 425.8125 ii X(O2)= 0.1 dX(NO)/dt= 3.431283E-06 X(N2)= 0.67 T= 2000 P= 1 k1= 425.8125 (𝒅[𝑵𝑶])/𝒅𝒕=𝟐𝒌_𝟏𝒇 [𝑶] _𝒆 [𝑵_𝟐] _𝒆
iii X(O2)= 0.05 dX(NO)/dt= 1.715642E-06 X(N2)= 0.67 T= 2000 P= 1 k1= 425.8125 iv X(O2)= 0.2 dX(NO)/dt= 9.234219E-07 X(N2)= 0.67 T= 1800 P= 1 k1= 51.56732 v X(O2)= 0.2 X(N2)= 0.67 dX(NO)/dt= 7.55527E-07 T= 2200 P= 1 k1= 51.56732
Part a.) Part b.) At steady-state, Part c.) Part d.) Calculate time to form NO Conditions: T 2100 0.167 MW kg/kmol 28.778 X(O,eq) 7.6E-05 X(O2,eq) 0.003025 X(N2,eq) 0.726 With equation 4.40, We can find the time using the following equation and equilibrium values, The equilibrium values for the reactants can be used and the integration w

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